Compressed-Air Discharge-Pressure Setpoint Savings

Because compression power rises with the pressure ratio, the isentropic power ratio between two setpoints gives the exact percent saved - which reproduces the DOE rule of about half a percent of compressor energy per psi. Dropping 120 to 105 psig on a 50 kW compressor saves ~7 percent, roughly $2,000 a year, for turning a knob. The reduced setpoint must still hold the minimum tool pressure after system drop, which is why fixing leaks and piping is what unlocks it. An energy-savings estimate, not a metered result.

work_current = ((current_psig + 14.7) / inlet_psia)^((k-1)/k) - 1; work_reduced = ((reduced_psig + 14.7) / inlet_psia)^((k-1)/k) - 1; pct_saved = 1 - work_reduced / work_current; kw_saved = input_kw x pct_saved; annual_kwh = kw_saved x run_hours; annual_savings = annual_kwh x rate_kwh. (k = 1.4)

The isentropic compression-power ratio (percent saved = 1 - [(P_reduced/P1)^((k-1)/k) - 1] / [(P_current/P1)^((k-1)/k) - 1]), which reproduces the DOE rule of roughly 0.5% of compressor energy per psi of reduction, by name; k = 1.4 for air is a named constant.

The isentropic pressure-ratio relation and the DOE per-psi rule are public; the DOE compressed-air tip sheets are free at energy.gov.

Estimate. AHJ and licensed professional govern.

Field names used by the API: current_psig, reduced_psig, inlet_psia, input_kw, run_hours, rate_kwh, pct_saved, kw_saved, annual_kwh, annual_savings

• Power ratio the saving is the isentropic work ratio between the two discharge pressures (absolute)isentropic compression
• Per-psi rule reproduces the DOE rule of about 0.5% of compressor energy per psi of reductionUS DOE
• Minimum pressure the reduced setpoint must still serve the tools after system pressure dropcompressed-air best practice